criterion of surjectivityCriterionOfSurjectivity2008-05-25 07:13:482008-05-25 07:13:48Theoremsurjectivesurjectivesurjection% this is the default PlanetMath preamble. as your knowledge
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\textbf{Theorem.}\, For surjectivity of a mapping \, $f\!:\,A \to B$,\, it's necessary and sufficient that
\begin{align}
B\!\smallsetminus\!f(X) \,\subseteq\, f(A\!\smallsetminus\!X) \quad \forall\, X \subseteq A.
\end{align}
{\em Proof.}\; $1^{\underline{o}}$.\, Suppose that\, $f\!:\,A \to B$\, is surjective.\, Let $X$ be an arbitrary subset of $A$ and $y$ any element of the set $B\!\smallsetminus\!f(X)$.\, By the surjectivity, there is an $x$ in $A$ such that\, $f(x) = y$, and since\, $y \notin f(X)$,\, the element $x$ is not in $X$, i.e.\, $x \in A\!\smallsetminus\!X$\, and thus\, $y = f(x) \in f(A\!\smallsetminus\!X)$.\, One can conclude that\, $B\!\smallsetminus\!f(X) \,\subseteq\, f(A\!\smallsetminus\!X)$\, for all\, $X \subseteq A$.
$2^{\underline{o}}$.\, Conversely, suppose the condition (1).\, Let again $X$ be an arbitrary subset of $A$ and $y$ any element of $B$.\, We have two possibilities:\\
a) $y \notin f(X)$; then\, $y \in B\!\smallsetminus\!f(X)$, and by (1), $y \in f(A\!\smallsetminus\!X)$.\, This means that there exists an element $x$ of\, $A\!\smallsetminus\!X \subseteq A$\, such that\, $f(x) = y$.\\
b) $y \in f(X)$; then there exists an $x \in X \subseteq A$\, such that\, $f(x) = y$.\\
The both cases show the surjectivity of $f$.